Since \(\mathcal {R}\)  exploits \(\mathcal {S}\)  or is fair relative to \(\mathcal {S}\) , we have \(a_{\mathcal {R}\mathcal {S}}\geqslant a_{\mathcal {S}\mathcal {R}}\) . Moreover, since \(a_{\mathcal {R}\mathcal {R}}\)  is maximal and the feasible region is convex, we must have \(a_{\mathcal {R}\mathcal {R}}\geqslant a_{\mathcal {S}\mathcal {R}}\) . If \(a_{\mathcal {R}\mathcal {R}}>a_{\mathcal {S}\mathcal {R}}\) , then \(\mathcal {R}\)  is automatically evolutionarily stable relative to \(\mathcal {S}\) . But if \(a_{\mathcal {R}\mathcal {R}}=a_{\mathcal {S}\mathcal {R}}\) , then \(a_{\mathcal {R}\mathcal {S}}\geqslant a_{\mathcal {S}\mathcal {R}}=a_{\mathcal {R}\mathcal {R}}\geqslant a_{\mathcal {S}\mathcal {S}}\) . If these inequalities are all equalities, then \(\mathcal {R}\)  is neutral relative to \(\mathcal {S}\) ; otherwise, \(\mathcal {R}\)  satisfies Maynard Smith's second condition [1]} and is thus evolutionarily stable relative to \(\mathcal {S}\) .

For example, in the battle of the sexes game, we know that FMTL is both fair and optimal against itself after sufficiently many learning steps (Fig. REF b). Thus, selfish learning cannot be evolutionary stable relative to FMTL. In this case, since selfish learning is also optimal against itself, FMTL is neutral relative to selfish learning. Of course, one can always deduce evolutionary stability by looking at the mean payoffs \(a_{\mathcal {X}\mathcal {Y}}\)  themselves. Proposition  REF simply says that a learner who performs optimally against itself is at least neutral against any nonexploitative opponent.
